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[SOLVED] Canonical Isomorphism and Tensor Products

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hi everyone, :)

Here's a problem that I have trouble understanding. Specifically I am not quite getting what it means by the expression \(\alpha (t)(v)\). Hope somebody can help me to improve my understanding. :)

Problem:

Let \(\alpha\) be the canonical isomorphism from \(V^*\otimes V\) to \(L(V,\, V)\). Find \(\alpha(t)(v)\) where \(t=(e^1+e^2)\otimes (e_3+e_4)\) and \(v=2e_1+3e_2+2e_3+3e_4\).
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Hi everyone, :)

Here's a problem that I have trouble understanding. Specifically I am not quite getting what it means by the expression \(\alpha (t)(v)\). Hope somebody can help me to improve my understanding. :)

Problem:

Let \(\alpha\) be the canonical isomorphism from \(V^*\otimes V\) to \(L(V,\, V)\). Find \(\alpha(t)(v)\) where \(t=(e^1+e^2)\otimes (e_3+e_4)\) and \(v=2e_1+3e_2+2e_3+3e_4\).
With the usual caveat that I'm not really at home with this tensor notation, the idea is that an element of $V^*\otimes V$ gives rise to a linear transformation from $V$ to $V$. The elementary tensor $e^i\otimes e_j$ gives rise to the linear transformation $T = \alpha(e^i\otimes e_j)$ defined by $T(v) = \alpha(e^i\otimes e_j)(v) = \langle e^i,v\rangle e_j$ (for all $v\in V$), where the angled brackets $\langle x,v\rangle$ denote the action of $x\in V^*$ on the element $x\in V$ under the duality between the two spaces. Thus $\alpha(e^i\otimes e_j)(e_k) = \begin{cases}e_j& \text{ if }k=i, \\ 0&\text{ if }k\ne i. \end{cases}$
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
With the usual caveat that I'm not really at home with this tensor notation, the idea is that an element of $V^*\otimes V$ gives rise to a linear transformation from $V$ to $V$. The elementary tensor $e^i\otimes e_j$ gives rise to the linear transformation $T = \alpha(e^i\otimes e_j)$ defined by $T(v) = \alpha(e^i\otimes e_j)(v) = \langle e^i,v\rangle e_j$ (for all $v\in V$), where the angled brackets $\langle x,v\rangle$ denote the action of $x\in V^*$ on the element $x\in V$ under the duality between the two spaces. Thus $\alpha(e^i\otimes e_j)(e_k) = \begin{cases}e_j& \text{ if }k=i, \\ 0&\text{ if }k\ne i. \end{cases}$
Thanks so much for your reply. I took a long time to understand this due to my limited knowledge about tensors, but I think now I am getting there. :)
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
As I indicated in another thread the basis of elementary tensors $e^j \otimes e_i$ of $V^{\ast} \otimes V$ can be identified with the basis of elementary matrices $E_{ij}$ of $\text{Hom}(V,V)$.

Note that this identification uses a basis choice for $V$, but it is possible to do this in a completely basis-free manner.

One thing to remember is that linear functionals (at least for finite-dimensional vector spaces) are pretty much just "glorified inner products". That is, every element $f \in V^{\ast}$ can be thought of as the function:

$\langle u,\_\rangle$ for some vector $u$.

For example, we have $e^j = \langle e_j,\_ \rangle$ for the standard basis for $V$.

***Note*** Vector spaces, of course, do not always come equipped with a "natural" inner product. However, any non-degenerate bilinear form $B$ can be used to induce an isomorphism (only in the finite-dimensional case, n.b.) between $V$ and $V^{\ast}$, and the "go-to" non-degenerate bilinear form in an inner product space is, of course, the inner product (BUT...in complex vector spaces, it is often more convenient to use a sesquilinear form, due to the peculiarities of the complex conjugate).

In particular, it is natural to identify the 1-form (dual basis element) $e^j$ with the $j$-th projection function.

*******

Computing the values of these mappings is often tedious. $\alpha(e^i \otimes e_j) (v)$ basically takes the $i$-th coordinate of $v$ (in the given "$e$" basis) and sticks it in the $j$-th slot, with every other coordinate 0. To extend this to the entire tensor $t$ we extend by (multi-, in this case, bi-) linearity.

So if:

$t = (e^1 + e^2) \otimes (e_3 + e_4) = e^1 \otimes e_3 + e^1 \otimes e_4 + e^2 \otimes e_3 + e^2 \otimes e_4$

Then:

$\alpha(t)(v) = 2e_3 + 2e_4 + 3e_3 + 3e_4 = 5e_3 + 5e_4$

or, perhaps more understandably, relative to our given basis $S$ of $V$:

$[T]_S([v]_S) = [T]_S(2,3,2,3)^T = (0,0,5,5)^T$

(if I have done my arithmetic correctly), that is relative to the given basis, our linear mapping $T$ has the matrix:

$\begin{bmatrix}0&0&0&0\\0&0&0&0\\1&1&0&0\\1&1&0&0 \end{bmatrix}$
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
As I indicated in another thread the basis of elementary tensors $e^j \otimes e_i$ of $V^{\ast} \otimes V$ can be identified with the basis of elementary matrices $E_{ij}$ of $\text{Hom}(V,V)$.

Note that this identification uses a basis choice for $V$, but it is possible to do this in a completely basis-free manner.

One thing to remember is that linear functionals (at least for finite-dimensional vector spaces) are pretty much just "glorified inner products". That is, every element $f \in V^{\ast}$ can be thought of as the function:

$\langle u,\_\rangle$ for some vector $u$.

For example, we have $e^j = \langle e_j,\_ \rangle$ for the standard basis for $V$.

***Note*** Vector spaces, of course, do not always come equipped with a "natural" inner product. However, any non-degenerate bilinear form $B$ can be used to induce an isomorphism (only in the finite-dimensional case, n.b.) between $V$ and $V^{\ast}$, and the "go-to" non-degenerate bilinear form in an inner product space is, of course, the inner product (BUT...in complex vector spaces, it is often more convenient to use a sesquilinear form, due to the peculiarities of the complex conjugate).

In particular, it is natural to identify the 1-form (dual basis element) $e^j$ with the $j$-th projection function.

*******

Computing the values of these mappings is often tedious. $\alpha(e^i \otimes e_j) (v)$ basically takes the $i$-th coordinate of $v$ (in the given "$e$" basis) and sticks it in the $j$-th slot, with every other coordinate 0. To extend this to the entire tensor $t$ we extend by (multi-, in this case, bi-) linearity.

So if:

$t = (e^1 + e^2) \otimes (e_3 + e_4) = e^1 \otimes e_3 + e^1 \otimes e_4 + e^2 \otimes e_3 + e^2 \otimes e_4$

Then:

$\alpha(t)(v) = 2e_3 + 2e_4 + 3e_3 + 3e_4 = 5e_3 + 5e_4$

or, perhaps more understandably, relative to our given basis $S$ of $V$:

$[T]_S([v]_S) = [T]_S(2,3,2,3)^T = (0,0,5,5)^T$

(if I have done my arithmetic correctly), that is relative to the given basis, our linear mapping $T$ has the matrix:

$\begin{bmatrix}0&0&0&0\\0&0&0&0\\1&1&0&0\\1&1&0&0 \end{bmatrix}$
Thanks so much. I will read all the details slowly to grab hold of them. The problem this tensor mathematics is that I find a lot of different approaches to a given problem with sometimes confuses me. :)